In a conductivity cell, the two platinum electrodes, each of area 10 cm^2 are fixed 1.5 cm apart. Th — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Determine the Effective Electrode Area ($A$) In a conductivity cell, the cell geometry is defined by the distance between the electrodes ($l$) and the cross-sectional area of the electrodes ($A$) in contact with the electrolyte: * The distance between the platinum electrodes ($l$) is $1.5\text{ cm}$. * The total physical area of each electrode is $10\text{ cm}^2$. Since the electrodes are **just half dipped** into the salt solution, only half of their total surface area is in contact with the electrolyte. Therefore, the effective area ($A$) to be used in calculations is: $$A = \frac{\text{Total Area}}{2} = \frac{10\text{ cm}^2}{2} = 5\text{ cm}^2$$ Step 2 - Calculate the Cell Constant ($G^*$) The cell constant ($G^*$) of a conductivity cell is defined as the ratio of the distance between the electrodes ($l$) to their effective surface area ($A$): $$G^* = \frac{l}{A}$$ Substituting the values determined in Step 1: $$G^* = \frac{1.5\text{ cm}}{5\text{ cm}^2} = 0.3\text{ cm}^{-1}$$ Step 3 - Calculate the Specific Conductance ($\kappa$) The specific conductance (or conductivity, $\kappa$) is related to the cell constant ($G^*$) and the electrical resistance ($R$) of the electrolyte solution by the formula: $$\kappa = \frac{G^*}{R}$$ We are given: * Resistance ($R$) = $50\ \Omega$ (or $50\text{ ohm}$) Substitute these values: $$\kappa = \frac{0.3\text{ cm}^{-1}}{50\ \Omega} = 0.006\ \Omega^{-1}\text{ cm}^{-1}$$ Step 4 - Calculate the Equivalent Conductance ($\Lambda_{eq}$) of the Solution The equivalent conductance ($\Lambda_{eq}$) of an electrolyte solution is related to its specific conductance ($\kappa$) and its normality ($N$) by the following formula: $$\Lambda_{eq} = \frac{\kappa \times 1000}{N}$$ We are given: * Normality of the salt solution ($N$) = $0.05\text{ N}$ (which is $0.05\text{ eq L}^{-1}$) Substituting the specific conductance ($\kappa = 0.006\ \Omega^{-1}\text{ cm}^{-1}$) and normality into the formula: $$\Lambda_{eq} = \frac{0.006\ \Omega^{-1}\text{ cm}^{-1} \times 1000\text{ cm}^3\text{ L}^{-1}}{0.05\text{ eq L}^{-1}}$$ $$\Lambda_{eq} = \frac{6\ \Omega^{-1}\text{ cm}^{-1}\text{ cm}^3\text{ L}^{-1}}{0.05\text{ eq L}^{-1}}$$ $$\Lambda_{eq} = \frac{6}{0.05}\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$$ $$\Lambda_{eq} = \boxed{120\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is correct:** As calculated, the equivalent conductance of the salt solution under the given conditions is exactly $120\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$. * **Option (B) is incorrect:** This value ($60$) is obtained if one forgets that the electrodes are half-dipped and incorrectly uses the total area ($A = 10\text{ cm}^2$) instead of the effective area ($A = 5\text{ cm}^2$), leading to an incorrect cell constant $G^* = 0.15\text{ cm}^{-1}$. * **Option (C) is incorrect:** This value ($240$) represents a mathematical error where the halving of the area is incorrectly applied as a multiplication, doubling the cell constant. * **Option (D) is incorrect:** This value ($3000$) is mathematically incorrect and does not satisfy the equivalent conductance formula. $$\text{Correct Option: } \boxed{\text{A}}$$